**SQUARE A 3 DIGIT NUMBER WHICH ARE NEAR THE NO WHOSE LAST TWO DIGITS ARE 00 (2)

if the given no is more 
RULE:-
we have to find (ABC)^2
first, subtract ABC - A*100= BC
the first part of the result will be (ABC+BC)*A
and the last two digits will be square of the subtraction result
if the square is more than two digit then put last two digits at last and add the extra with first part

.................................................................................................................
EX:-
508^2=?
508-500=8
the first part of the result will be (508+8)*5=516*5=2580
and the last two digits will be 8^2=64
so the result is 258064

**SQUARE A 3 DIGIT NUMBER WHICH ARE NEAR THE NO WHOSE LAST TWO DIGITS ARE 00 (1)

if the given no is less 
RULE:-
we have to find (ABC)^2
first, subtract that no from it's next nearest no whose last two digits are 0
suppose it becomes X
(  that means X=[(A+1)*100-ABC]  )
then, first part of the result will be (ABC - X)*(A+1)
and last two digits of the result will be X^2
if the square is more than two digit then put last two digits at last and add the extra with first part

......................................................................................
EX:-
(298)^2=?
now it's next nearest no is 300
so 300-298=2
so first part will be =(298-2)*(2+1)=296*3=888
and last two digits will be 2^2= 04
so ans is 88804

482^2=?
it's next nearest no is 500
so 500-482=18
so first part will be= (482-18)*(4+1)=464*5=2320
and the last two digits will be 18^2=3 24
so the ans is 232 (0+3) 24=232324

** SOME DIVISION RULES FOR REPEATION OF THE NUMBERS

if a number is repeated 2 or its multiplied no of times then it will be divided by 11
if a number is repeated 6 or its multiplied no of times then it will be divided by 3/7/11/13/37
if two number is repeated 3 or its multiplied no of times then it will be divided by 3/7/13/37
if 0 is repeated at-least 2 times at the last then it will be divided by 4
if 0 is repeated at-least 3 times at the last then it will be divided by 8

(we know the every rules but when the combination is used then we forgot about these formulas)

.............................................................................................................................
EX:-

#) 44444....4 here 4 is repeated 429 times.if it is divided by 1144.then what will be the remainder?
Ans:- here 1144=8*11*13
now 4444....4 if it is repeated 426 times and the last three digits are 000
then this will be divided by 1144 or (11*13*8) as 426 divided by 6 and as last three digits are 0 so it will be divided by 8
so remainder is 444

#) 1212...12 here 12 is repeated 283 times.if it is divided by 1036. then what will be the remainder?
Ans:- here 1036= 4*7*37
now 1212...12 if it is repeated 282 times and the last two digits are 00
then this will be divided by 1036 or (4*7*37) as 282 is divided by 3 and as last two digits are 0 so it will be divided by 4
so remainder is 12

**SQUARE A NUMBER WHICH ARE NEAR 100(2)

if the given no is more than 100 
first take that no
now first subtract 100 from that no
the first part of the result will be (the given no+the subtraction result )
and the last two digits will be square of the subtraction result
if the square is more than two digit then put last two digits at last and add the extra with first part


.......................................................................................
EX:-
108^2=?
108-100=8

here first part is 108+8=116
and last two digit will be 8^2=64
so the ans will be=11664

116^2=?
116-100=16
here first part is 116+16=132
and last two digit will be 16^2=256
132+2=134
so the ans will be=13456

**SQUARE A NUMBER WHICH ARE NEAR 100(1)

if the given no is less than 100 
first take that no
now first subtract that number from 100
the first part of the result will be (the given no-the subtraction result )
and the last two digits will be square of the subtraction result
if the square is more than two digit then put last two digits at last and add the extra with first part

.......................................................................................
EX:-
97^2=?
100-97=3
so first part will be 97-3= 94
and last two digit will be 3^2=09
so the ans will be 9409

84^2=?
100-84=16
so the first part will be 84-16=68
and the last two digit will be 16^2= 256
68+2=70
so the ans will be 7056

**MULTIPLY TWO NUMBERS WHICH ARE NEAR 100(2)

if the given nos are more than 100
take that two nos---
at first subtract 100 from those number
now, the first three digit of multiplication result will be (any given no + another's subtraction result )
and the last two digit will be multiplication of those two subtraction result

...........................................................................................
EX:-
104*103=?
here 104 is 4 more than 100 and 103 is 3 more than 100
so the first three digit will be (104+3) or (103+4) or 107
and last two digit will be 3*4 or 12
so the ans will be 10712

**MULTIPLY TWO NUMBERS WHICH ARE NEAR 100(1)

if the given nos are less than 100
take that two nos---
at first subtract those numbers from 100
now, the first two digit of multiplication result will be (any given no - another's subtraction result )
and the last two digit will be multiplication of those two subtraction result

...............................................................................................................................
EX:-
92*94=?
here 92 is 8 less than 100 and 94 is 6 less than 100
so, first two digit will be (92-6) or (94-8) or 86
and last two digit will be 6*8=48
so the ans will be 8648

**FOR MAKING SQUARE ROOT OF A NUMBER

this formula works when the main no is in between 10 to 99

for this we have to remember only ----

if the last digit of the given no is 0 then square root's last digit will be 0
if the last digit of the given no is 1 then square root's last digit will be 1 / 9
if the last digit of the given no is 4 then square root's last digit will be 2 / 8
if the last digit of the given no is 9 then square root's last digit will be 3 / 7 
if the last digit of the given no is 6 then square root's last digit will be 4 / 6
if the last digit of the given no is 5 then square root's last digit will be 5

and
1*2=2
2*3=6
3*4=12
4*5=20
5*6=30
6*7=42
7*8=56
8*9=72

RULE:-
if the given no is (ABCD) and its square root is (XY)      (A may be 0 or any digit)
then AB is near to (X*(X+1))
and Y can be determined by seeing D
if AB is just less than (X*(X+1)), then Y will be the lesser no
and
if AB is just bigger than (X*(X+1)), then Y will be the bigger no
 ...................................................................................................

EX:-
3969 is given
now 39 is just less than 6*7
so first digit of it's square root is 6
and as last digit of the given no is 9
so the last digit of it's square root will be either 3 or 7
now as 39 is less than 6*7
so last digit will be 3
so the ans is 63


784 is given
now 7 is just bigger than 2*3
so first digit of it's square root is2
and as last digit of the given no is 4
so the last digit of it's square root will be either 2 or 8
now as 39 is bigger than 6*7
so last digit will be 8
so the ans is 28

**FOR MAKING CUBE ROOT OF A NUMBER

this formula works when the main no is in between 10 to 99

we know the cubes are as follows
1^3=1
2^3=8
3^3=27
4^3=64
5^3=125
6^3=216
7^3=343
8^3=512
9^3=729

now the cubes of 10 to 99 will be in between 1000 to 1000000
so when a no is given just take the last three digit as different and the rest of first digits as different
from the last digit of last three digit we get the last number of it's cube root
now or first digit see the first part... the fist part is greater than whose cube take that no as first digit

if we go through the example then we understand it clearly

....................................................................
EX:-
     46656= 46 656
here the last three digits 656. here the last digit is 6 which is in 6's cubes last digit
so the last digit is 6
now the rest first digits is 46
which is in just greater than 27 that means 3's cube
so, the cube root of the no 46656 is 36

**FOR SQUARE OF THE NO ENDNG WITH NINE(9)

leave 9 , now multiply the (first part) with (first part+1). put the result in first
then add (8* first part) with the first digit of (9^2 or 81) that means with 8.if here an excess is present then add that with the result of multiplication.
and at last put 1

..................................................................................

EX:-
19^2= (1*2) (8*1+8) 1 = (2+1) 6 1 = 361
119^2= (11*12) (8*11+8) 1 =(132+9) 6 1 = 14161

**FOR SQUARE OF THE NO ENDNG WITH EIGHT(8)

leave 8 , now multiply the (first part) with (first part+1). put the result in first
then add (6* first part) with the first digit of (8^2 or 64) that means with 6.if here an excess is present then add that with the result of multiplication.
and at last put 4

...................................................

EX:-
38^2= (3*4) (3*6+6) 4 = (12+2) 4 4 = 1444
128^2=(12*13) (12*6+6) 4 = (156+7) 8 4 =16384

**FOR SQUARE OF THE NO ENDNG WITH SEVEN(7)

leave 7 , now multiply the (first part) with (first part+1). put the result in first
then add (4* first part) with the first digit of (7^2 or 49) that means with 4.if here an excess is present then add that with the result of multiplication.
and at last put 9

...................................................

EX:-
37^2= (4*3) (4*3+4) 9 = (12+1) 6 9 = 1369
117^2= (11*12) (4*11+4) 9 = (132+4) 8 9 = 13689

**FOR SQUARE OF THE NO ENDNG WITH SIX(6)

leave 6 , now multiply the (first part) with (first part+1). put the result in first
then add (2* first part) with the first digit of (6^2 or 36) that means with 3.if here an excess is present then add that with the result of multiplication.
and at last put 6

...................................................

EX:-
26^2= (2*3) (2*2+3) 6 = 676
46^2= (4*5) (4*2+3) 6 = (20+1) 1 6 = 2116

**FOR SQUARE OF THE NO ENDNG WITH FIVE(5)

leave 5 , now multiply the first part of that no with (first part+1). put 5^2 or 25 after the result of multiplication

......................................................................
EX:-
45^2= (4*5) 25=2025
265^2= (26*27) 25=70225

**FOR SQUARE OF THE NO ENDNG WITH FOUR(4)

at first square the last digit of the given no and put it in the last and here the square becomes 16 so put 6 here and add 1 with first part

then in first put ((first part^2)*10+(first part*last part*2)).
then the first part becomes  ((first part^2)*10+(first part*last part*2)+1).
.....................................................................................................

EX:-
44^2= ((4^2)*10+(4*4*2)+1) 6=(160+32+1) 6=1936

**FOR SQUARE OF THE NO ENDNG WITH ONE OR TWO OR THREE (1/2/3)

at first square the last digit of the given no and put it in the last
then in first put ((first part^2)*10+(first part*last part*2)).

...............................................................................

EX:-
11^2=((1^2)*10+(1*1*2)) 1^2 = (10+2) 1=121
131^2=((13^2)*10+(13*1*2)) 1^2=(1690+26) 1= 17161

22^2= ((2^2)*10+(2*2*2)) 2^2=(40+8) 4=484

33^2=((3^2)*10+(3*3*2)) 3^2=(90+18) 9=1089

**MULTIPLY TWO 3 DIGIT NUMBERS MENTALLY

the name of the trick is the cross multiplication technique

lets start with the numbers...  987*654

Step 1--- arrange the numbers in order (one on top of the other) your pick.

987
654
----------

start by first multiplying 7*4=28. write down the 8 and carry the  2 mentally.

987
654
---------- CARRY 2
----8

Step 2---- attempt to form an invisible X
It may not sound simple at first but I'll explain.

Start by observing the 8 & 4, then the 5 & 7. Make a line between those numbers and you will get an X.

Now multiple 8*4=32 PLUS(+) 5*7=35
so 32+35=67
67 + 2 (the carry)= 69
Now write down the 9 and carry the 6 mentally.

987
654
---------- CARRY 6
---98

Step 3--- now examine the numbers 4 & 9, 6 & 7, then in the middle 8 & 5.

Multiply them and add em' all up:
4*9=36  +6*7=42  +8*5= 40
so 36+42+40 = 118 PLUS carry 6 = 124.

Write down the 4 and carry the 12.

987
654
---------- CARRY 12
--498

Step 4--- Now we will attempt the invisible X for the last time. This time take notice of the numbers 9 & 5, 6 & 8
multiply them and add em' all up:
5*9=45 + 6*8=48
so 45 +48 = 93 PLUS carry 12 = 105.
Write down the 5 and carry the 10.

987
654
---------- CARRY 10
--5498

Step 5 (final step)--- In the beginning we started by multiplying 7 & 4. Now we will multiply 9 & 6.
so 9*6=54 + carry 10 = 64

Write it down and the behold the FINAL ANSWER!

987
654
----------
645498

 

 

**SQUARE OF ANY 2 DIGIT NUMBER

explain this rule by taking examples
27^2 = (27+3)*(27-3) + 3^2 = 30*24 + 9 = 720+9 = 729
In this method, we have to make a number ending with 0, that's why; we add 3 to 27

..................................................................................
EX:-
78^2 = (78+2)*(78-2) + 2^2 = 80*76 + 4 = 6080+4 = 6084

**MULTIPLICATION OF 11 WITH ANY 3 DIGIT NUMBER

explain this rule by taking examples
1. 352*11 = 3---(3+5)---(5+2)---2 = 3872
Means insert the sum of first and second digits, then sum of second and third digits between the two terminal digits of the number
2. 213*11 = 2---(2+1)---(1+3)---3 = 2343

if sum of two digits of the number is greater than 10, then add 1 to previous digit and subtract 10 to the associated digit.

..............................................................................
EX:-
1) 329*11 = 3--- (3+2) +1--- (2+9-10) ---9 = 3619
2) 758*11 = 7+1---(7+5-10)+1---(5+8-10)---8 = 8338

**MULTIPLICATION OF 11 WITH ANY 2 DIGIT NUMBER

explain this rule by taking examples
35*11 = 3---(3+5)---5 = 385
Means insert the sum of first and second digits between the two digits of the number
27*11 = 2---(2+7)---7 = 297


 if sum of two digits of the number is greater than 10, then add 1 to first digit and subtract 10 to the inserted term
......................................................................

EX:-
39*11 = 3+1---(3+9-10)---9 = 429

71*11=7+1----(7+1)---1=881

**MULTIPLICATION OF TWO NUMBERS , IN BETWEEN 11 TO 19

explain this rule by taking examples
13*19 = (13+9)*10 + (3*9) = 220 + 27 = 247
Means add first number and last digit of the second number take zero in the third place of this number then add product of last digit of the two numbers in it

....................................................................................


18*14 = (18+4)*10 + (8*4) = 220 + 32 = 252

**MULTIPLICATION OF ANY NUMBER BY 11

explain this rule by taking examples
1) 234163*11 = 2---2+3---3+4---4+1---1+6---6+3---3 = 2575793
Means insert the sum of 2 successive digits and put 2 terminal digits in its place
2) 45345181*11 = 4---4+5---5+3---3+4---4+5---5+1---1+8---8+1---1 = 498796991

 if sum of two digits of the number is greater than 10, then add 1 to first digit and subtract 10 to the inserted term
...................................................................

374827*11 = 3+1--- (3+7-10) +1--- (7+4-10) +1--- (4+8-10) +1--- (8+2-10)--- (2+7) ---7 = 4123097